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Topic: 36 BALLS, one different (Read 1114 times) |
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pcbouhid
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36 BALLS, one different
« on: Nov 18th, 2005, 5:35am » |
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I think that everyone knows the problem of the "12 balls", one different (lighter or heavier than the others), 3 weighings.
What about 36 balls, one different (lighter or heavier), 4 weighings?
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Grimbal
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Re: 36 BALLS, one different
« Reply #1 on: Nov 18th, 2005, 7:36am » |
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I think you can even go up to 39.
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Joe Fendel
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Re: 36 BALLS, one different
« Reply #2 on: Nov 18th, 2005, 7:44am » |
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Use 1 weighing to reduce to the 12-ball problem.
Weigh balls 1-12 against 13-24. If one side or the other is heavier, this can now be reduced to the 12-ball problem by considering the 12 pairs: 1-13, 2-14, ... 12-24. One of these pairs of balls is heavier or lighter. Use three weighings to figure out which pair and whether it's light or heavy - that will be sufficient information to determine which ball in the pair is off-weight and in which direction. (For example, if 1-12 is heavier than 13-24 in weighing 1, and you learn in the subsequent 3 weighings that the 1-13 pair is heavy, then you know that in fact ball 1 is heavy.)
If the first weighing is balanced, then we're reduced to the 12-ball problem on balls 25-36.
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Grimbal
wu::riddles Moderator
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Re: 36 BALLS, one different
« Reply #3 on: Nov 18th, 2005, 7:57am » |
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For 36 balls, it is actually easier.
Make 12 groups of 3, use the usual algorthm to find the ligher/heavier group and use the extra weighing to find which one of the 3 balls in the identified group is the different one.
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Neelesh
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Re: 36 BALLS, one different
« Reply #4 on: Nov 18th, 2005, 8:13am » |
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on Nov 18th, 2005, 7:36am, Grimbal wrote:| I think you can even go up to 39. |
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Yes, true.
(3w-3)/2 is the maximum number of balls that can be weighed with w weighings
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